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Math Help - Difficult Chain Rule Problem

  1. #1
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    Difficult Chain Rule Problem

    I believe it uses the chain rule, anyway. My professor told me in class the other day that the chain rule can be used so I would prefer if there was an answer utilizing the rule, but really, anything will do. These webwork problems kill me! TIA!

    If ,

    Calculate .

    help!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by funnylookinkid View Post
    I believe it uses the chain rule, anyway. My professor told me in class the other day that the chain rule can be used so I would prefer if there was an answer utilizing the rule, but really, anything will do. These webwork problems kill me! TIA!

    If ,

    Calculate .

    help!
    By the chain rule (look this up): \frac d{dx} (f(3x^2)) = f'(3x^2) \cdot 6x

    we are told that this is equal to 7x^4

    thus we have f'(3x^2) \cdot 6x = 7x^4

    \Rightarrow f'(3x^2) = \frac 76x^3

    i leave the rest to you
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  3. #3
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    I get all of that and I get that dy/dx = dy/du * du/dx and I have what you have above. I just cannot figure out how to relate the 3x^2 to x and what to do to differentiate at x. I'm trying to use all of the classic formulas of the chain rule but I have found myself quite baffled! I'm not fishing for an answer here, but maybe there's something I'm missing?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by funnylookinkid View Post
    I get all of that and I get that dy/dx = dy/du * du/dx and I have what you have above. I just cannot figure out how to relate the 3x^2 to x and what to do to differentiate at x. I'm trying to use all of the classic formulas of the chain rule but I have found myself quite baffled! I'm not fishing for an answer here, but maybe there's something I'm missing?
    there is no more calculus left to be done. the rest is basically using your knowledge of composite functions. what does f'(3x^2) mean? this is a composite function. it means you take f'(x) and replace all the x's you see in it with 3x^2. just work backwards.
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  5. #5
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    I arrived at (7/18)x and it is not registering as a correct answer. Is that what I should have arrived at?
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    I figured since you were plugging in 3x^2 for all xs, I set up the following:

    3x^2 * (t) = (7/6)x^3, where t = what I had to multiply 3x^2 by in order to reach my final answer. The answer I got was (7/18)x. I guess I'm missing something?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by funnylookinkid View Post
    I figured since you were plugging in 3x^2 for all xs, I set up the following:

    3x^2 * (t) = (7/6)x^3, where t = what I had to multiply 3x^2 by in order to reach my final answer. The answer I got was (7/18)x. I guess I'm missing something?
    ok, so the reason you are having trouble with this is that you do not have a good grasp of composite functions. making composite function and multiplying functions are two totally different things! replacing x with a function is not the same as multiplying by the function.

    if you have f(x) = x^2

    then f(x + 2) = (x + 2)^2 NOT x^2(x + 2)

    now try to work from that idea. you want to somehow write the function in terms of 3x^2, and then replace the 3x^2 with x. that is all
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  8. #8
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    Okay. I think I understand now. Here's what I did now:

    I rewrote the function of (7/6)x^3 as 3x^2 * (7/18)x, since I'm trying to rewrite the derivative in terms of 3x^2 so I can replace it with just x. I did that and was left with x * (7/18)x, multiplied it out to get (7/18)x^2.

    But the answer is still wrong. I thought I had a pretty good grasp on this stuff, but apparently I was wrong! Am I still missing something?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by funnylookinkid View Post
    Okay. I think I understand now. Here's what I did now:

    I rewrote the function of (7/6)x^3 as 3x^2 * (7/18)x, since I'm trying to rewrite the derivative in terms of 3x^2 so I can replace it with just x. I did that and was left with x * (7/18)x, multiplied it out to get (7/18)x^2.

    But the answer is still wrong. I thought I had a pretty good grasp on this stuff, but apparently I was wrong! Am I still missing something?
    that is wrong. you have an x left over. the only variable you want in your function is 3x^2, you want no x's beside it or anything. it must only be 3x^2 times some constant. try once more (plus, you should have realized that if you plug in 3x^2 into (7/18)x^2 you do not get (7/6)x^3 as you should)
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  10. #10
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    I think I'm gonna have to sleep on it, haha. I'll be back in the morning, undoubtedly, hopefully with something more than I have now!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by funnylookinkid View Post
    I think I'm gonna have to sleep on it, haha. I'll be back in the morning, undoubtedly, hopefully with something more than I have now!
    note that \frac 76x^3 = \frac 76 \left(\frac {{\color{red}3x^2}}3 \right)^{3/2}

    how did i get that?

    i'm leaving, good luck
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  12. #12
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by Jhevon View Post
    By the chain rule (look this up): \frac d{dx} (f(3x^2)) = f'(3x^2) \cdot 6x

    we are told that this is equal to 7x^4

    thus we have f'(3x^2) \cdot 6x = 7x^4

    \Rightarrow f'(3x^2) = \frac 76x^3

    i leave the rest to you
    I like this guy's method, watch and learn

    YouTube - 1991 AHSME #21 - Functions
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    Quote Originally Posted by DivideBy0 View Post
    I like this guy's method, watch and learn

    YouTube - 1991 AHSME #21 - Functions
    I seen other stuff by this guy. The only problem with his videos is he talks way too fast. But that is probably because he needs to explain all of that in under 10 minute, for that is the boundary.
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